For a van der Waal's gas, the critical pressure and temperature are 73.89 atm and 27°C, respectively — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Volume occupied by the molecules themselves is $nb$, where $b = V_c/3$ or derived from critical constants: $b = \frac{R T_c}{8 P_c}$. $T_c = 27^\circ\text{C} = 300\text{ K}$, $P_c = 73.89\text{ atm}$. Using $R = 0.0821\text{ L-atm/K-mol}$, $b = \frac{0.0821 \times 300}{8 \times 73.89} = \frac{24.63}{591.12} \approx 0.04166\text{ L/mol}$. For $n=24\text{ moles}$, the excluded volume $nb = 24 \times 0.04166 \approx 1.0\text{ L}$. However, the volume occupied *only* by the molecules is their actual physical volume, which is $nb/4$. True volume $= 1.0\text{ L} / 4 = 0.25\text{ L} = 250\text{ ml}$. Formatted as a four-digit integer, it is 0250. Therefore, correct answer is 0250.